Finding f'(0) with L'Hopital's Rule and a Piecewise Function

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Homework Statement



Find f'(0) given a piecewise function defined as

f(x) =
{g(x)/x2, x≠0
{0, x=0

where g(x) is a function satisfying g(0)=g'(0)=g''(0) and g'''(0)=14


Homework Equations


none.

The Attempt at a Solution



So far, I've reasoned that for f to be differentiable at 0, limit as x approaches 0 of g(x)/x2 is zero, and that leads to L'Hopital's rule (?). However, I get stuck after 2 derivations with 0/2 where the rule no longer applies, and I'm unable to use the fact that g'''(x)=14.

Any help would be much appreciated :smile:
 
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I'm assuming that g(0)=0. You forgot to mention that.

It is indeed true that after 2 derivations you get 0/2. This is correct. This means that the limit is 0. Thus you have found that

[tex]\lim_{x\rightarrow 0}{\frac{g(x)}{x^2}}=0[/tex]

But this only means that f is continuous at 0. You need to calculate another limit to find that f is differentiable at 0.
 
micromass said:
I'm assuming that g(0)=0. You forgot to mention that.

It is indeed true that after 2 derivations you get 0/2. This is correct. This means that the limit is 0. Thus you have found that

[tex]\lim_{x\rightarrow 0}{\frac{g(x)}{x^2}}=0[/tex]

But this only means that f is continuous at 0. You need to calculate another limit to find that f is differentiable at 0.

Yeah, forgot to mention that sorry.

I actually took a different approach to the question while I was thinking about it in class. I figured that using L'Hopital's rule multiple times only seems to prove that the function is continuous.

I integrated g'''(x) multiple times to get an equation g(x)=7x3/3, making f(x)=7x/3 and then f'(0)=7/3. Is this valid? working backwards, everything seems to be consistent, only the piecewise part is just making me feel somewhat uneasy about doing it this way.
 
micromass said:
But you only know the value of g in 0. For all you know, the function will be something very different...

Wait, but g'''(0)=14

integrate that, g''(x)=14x+C, C=0 since g''(0)=0 through substitution

integrate, g'(x)=7x2+C, again C=0 through substitution

integrate, g(x)=7x3/3+C, C=0 through substitution

then plug that g(x) into the piecewise function to get f(x)=7x/3

Seems to work, doesn't it?
 
micromass said:
I'll give you a counterexample. The function

[tex]g(x)=\frac{14}{6}x^3e^x[/tex]

is also a good choice for g.

ahh, now I'm confused as f**, what did I do wrong in the integrals?

Also, the counterexample you gave does yield the same answer to f'(0)=7/3, and just out of curiosity, how did you come up with that counterexample so quickly, I'm quite baffled, are you a mathematician or prof or something?
 
FsLiu said:
Wait, but g'''(0)=14

integrate that, g''(x)=14x+C
No, you can't say this.

If it was given that g'''(x) = 14, then you could say that g''(x) = 14x + C, but you don't have any information about g''(x), only its value when x = 0.
FsLiu said:
, C=0 since g''(0)=0 through substitution

integrate, g'(x)=7x2+C, again C=0 through substitution

integrate, g(x)=7x3/3+C, C=0 through substitution

then plug that g(x) into the piecewise function to get f(x)=7x/3

Seems to work, doesn't it?
 
Mark44 said:
No, you can't say this.

If it was given that g'''(x) = 14, then you could say that g''(x) = 14x + C, but you don't have any information about g''(x), only its value when x = 0.

Wait but the original question stated that g(0)=g'(0)=g''(0)=0, sorry I forgot to edit that in.

I asked a second year student about it though, and he thinks it'd suffice because I found one general solution to the problem, and the question is only asking for f'(0) which I found to be 7/3.
 
It may not suffice, but what is true, assuming g(x)=7x3/3, is g'(0)=g''(0)=0 and g'''(0)=14.

And the question asks for the value of f'(0), so any possible equation of g(x) would suffice.

Thanks for all the input guys, but I'm going conclude that question there and ask the TA's/profs about it tomorrow. Brain hurts from this.